From: Ralph Shnelvar <ralphs@...32.com>
Date: 2012-02-24T00:53:57+09:00
Subject: Re: How are closures implemented?

Robert,

Thursday, February 23, 2012, 6:10:28 AM, you wrote:

RK> On Mon, Feb 20, 2012 at 11:23 PM, Brian Candler <b.candler@pobox.com> wrote:
>> Ralph Shnelvar wrote in post #1047769:
>>> How are closures implemented? �How does the closure know about its
>>> environment?
>>
>> The closure *is* the environment.
>>
>> Are you familiar with local variables allocated on a stack? Then instead
>> imagine stack frames allocated out of the heap, in a linked list.
>> Instead of being deallocated at the end of a function, they are
>> garbage-collected when there are no more references to them, just like
>> any other object.

RK> This is the best explanation about closures I have read in a long time, Brian!

RK> Ralph, try this and see whether it helps the explanation sink in:

RK> def f(x)
RK>   [
RK>     lambda { x += 1 },
RK>     lambda { x += 0.5 },
RK>   ]
RK> end

RK> a, b = f(0)

RK> 3.times do
RK>   p a.call, b.call
RK> end

RK> Note how there is just a single environment which is shared by the two closures.


Ok ... modifying your example


class Y
  z = 10
  
  define_method :f do |x|
    [
      lambda { z += z     ; puts("z += z:   z=#{z}, x=#{x}"); x },
      lambda { x += 0.5;  ; puts("x += 0.5: z=#{z}, x=#{x}"); x },
    ]
  end
end

a, b = Y.new.f(0)

3.times do
  puts("a: #{a.call}")
  puts("b: #{b.call}")
end



Is z "on the stack" or "merely" in scope?

As I understand it, z is bound to the closure.

But how does Ruby know?